Question: Solve for $x$ : $ 4|x - 3| - 5 = -2|x - 3| + 6 $
Solution: Add $ {2|x - 3|} $ to both sides: $ \begin{eqnarray} 4|x - 3| - 5 &=& -2|x - 3| + 6 \\ \\ { + 2|x - 3|} && { + 2|x - 3|} \\ \\ 6|x - 3| - 5 &=& 6 \end{eqnarray} $ Add ${5}$ to both sides: $ \begin{eqnarray} 6|x - 3| - 5 &=& 6 \\ \\ { + 5} &=& { + 5} \\ \\ 6|x - 3| &=& 11 \end{eqnarray} $ Divide both sides by ${6}$ $ \dfrac{6|x - 3|} {{6}} = \dfrac{11} {{6}} $ Simplify: $ |x - 3| = \dfrac{11}{6}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{11}{6} $ or $ x - 3 = \dfrac{11}{6} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{11}{6} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{11}{6} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{11}{6} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $6$ $ x = - \dfrac{11}{6} {+ \dfrac{18}{6}} $ $ x = \dfrac{7}{6} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{11}{6} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{11}{6} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{11}{6} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $6$ $ x = \dfrac{11}{6} {+ \dfrac{18}{6}} $ $ x = \dfrac{29}{6} $ Thus, the correct answer is $x = \dfrac{7}{6} $ or $x = \dfrac{29}{6} $.